Solve 2/x=2x-3

$$\frac { 2 } { x } = 2 x - 3$$

Solve for x

$$x=2$$

$$x=-\frac{1}{2}=-0.5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$2=2xx+x\left(-3\right)$$

Multiply x and x to get x^{2}.

$$2=2x^{2}+x\left(-3\right)$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+x\left(-3\right)=2$$

Subtract 2 from both sides.

$$2x^{2}+x\left(-3\right)-2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-3x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-2\right)}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-2\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\times 2}$$

Add 9 to 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-\left(-3\right)±5}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±5}{4}$$

Now solve the equation x=\frac{3±5}{4} when ± is plus. Add 3 to 5.

$$x=\frac{8}{4}$$

Divide 8 by 4.

$$x=2$$

Now solve the equation x=\frac{3±5}{4} when ± is minus. Subtract 5 from 3.

$$x=-\frac{2}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=2 x=-\frac{1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$2=2xx+x\left(-3\right)$$

Multiply x and x to get x^{2}.

$$2=2x^{2}+x\left(-3\right)$$

Swap sides so that all variable terms are on the left hand side.

$$2x^{2}+x\left(-3\right)=2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-3x=2$$

Divide both sides by 2.

$$\frac{2x^{2}-3x}{2}=\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{3}{2}x=\frac{2}{2}$$

Divide 2 by 2.

$$x^{2}-\frac{3}{2}x=1$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=1+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=1+\frac{9}{16}$$

Add 1 to \frac{9}{16}.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{5}{4} x-\frac{3}{4}=-\frac{5}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=2 x=-\frac{1}{2}$$